結果
問題 | No.1035 Color Box |
ユーザー |
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提出日時 | 2021-08-10 22:01:12 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 82 ms / 2,000 ms |
コード長 | 2,411 bytes |
コンパイル時間 | 970 ms |
コンパイル使用メモリ | 104,796 KB |
最終ジャッジ日時 | 2025-01-23 17:25:37 |
ジャッジサーバーID (参考情報) |
judge5 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 36 |
ソースコード
#include <algorithm>#include <cmath>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <set>#include <string>#include <tuple>#include <vector>#define mkp make_pair#define mkt make_tuple#define rep(i, n) for (int i = 0; i < (n); ++i)#define all(v) v.begin(), v.end()using namespace std;typedef long long ll;const ll MOD = 1e9 + 7;// const ll MOD = 998244353;template <class T>void chmin(T &a, const T &b) {if (a > b) a = b;}template <class T>void chmax(T &a, const T &b) {if (a < b) a = b;}#define MAX_N 1000010ll inv[MAX_N + 10], fac[MAX_N + 10], ifac[MAX_N + 10];void setComb() {inv[0] = 1;inv[1] = 1;fac[1] = 1;ifac[1] = 1;fac[0] = 1;ifac[0] = 1;for (int i = 2; i < MAX_N; i++) {inv[i] = (-MOD / i) * inv[MOD % i] % MOD;fac[i] = fac[i - 1] * i % MOD;ifac[i] = ifac[i - 1] * inv[i] % MOD;inv[i] = (inv[i] + MOD) % MOD;fac[i] = (fac[i] + MOD) % MOD;ifac[i] = (ifac[i] + MOD) % MOD;}return;}ll comb(ll n, ll k) {if (n < k || n < 0 || k < 0)return 0;elsereturn ((fac[n] * ifac[k] % MOD * ifac[n - k] % MOD + MOD) % MOD);}ll hcomb(ll n, ll r) { // this size is really ok??if (n == 0 && r == 0)return 1;else if (n < 0 || r < 0)return 0;elsereturn comb(n + r - 1, r);}ll binom(ll n, ll k) {if (n < k || n < 0 || k < 0) return 0;n %= MOD;ll res = 1;for (ll i = 0; i < k; i++) res = res * ((n - i + MOD) % MOD) % MOD;res = res * ifac[k] % MOD;return res;}ll mod_pow(ll x, ll n) {x %= MOD;ll res = 1;while (n > 0) {if (n & 1) res = res * x % MOD;x = x * x % MOD;n >>= 1;}return res;}ll mod_inverse(ll x) {return mod_pow(x, MOD - 2);}void add(ll &a, ll b) {a = (a + b) % MOD;}void mul(ll &a, ll b) {a %= MOD;b %= MOD;a = a * b % MOD;}void solve() {setComb();int N, M;cin >> N >> M;vector<ll> dp(M + 1, 0);for (int i = 1; i <= M; i++) {ll res = 1;mul(res, comb(M, i));mul(res, mod_pow(i, N));add(res, MOD - dp[i - 1]);dp[i] = res;}ll ans = dp[M];cout << ans << endl;}int main() {cin.tie(nullptr);ios::sync_with_stdio(false);solve();return 0;}